Question

The vertices of xyz are x(-3,-6), Y(21,-6), and z(21,4). What is the perimeter of the triangle

Answer 1

Note that this a right-angled triangle with the right angle at Y.

Use coordinates of the point to find XY and YZ

XY = 21 - (-3)

= 24

YZ = 4 - (-6)

= 10

Use Pythagoras theorem to XZ

XZ = sqrt[24^2 + 10^2]

= 26

Perimeter = 26 + 24 + 10

= 60

Answer 2

60 units

Explanation:

Given: The vertices of XYZ are
`X(-3,-6)`,
`Y(21,-6)`, and
`Z(21,4)`

To find: Perimeter of the triangle.

Solution:

We know that the distance between two points
`A(x_(1),y_(1))` and
`B(x_(2),y_(2))` is
`\sqrt{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2) }`

We have, the vertices of triangle XYZ as
`X(-3,-6)`,
`Y(21,-6)`, and
`Z(21,4)`

So,

`XY=\sqrt{(21+3)^(2)+(-6+6)^(2)}=24`

`YZ=\sqrt{(21-21)^(2)+(4+6)^(2)}=10`

`ZX=\sqrt{(21+3)^(2)+(4+6)^(2)`

`ZX=\sqrt{(24)^(2)+(10)^(2)`

`ZX=√(576+100)`

`ZX=√(676)=26`

Now, we have
`XY=24`,
`YZ=10`, and
`ZX=26`

Perimeter of Δ
`XYZ=24+10+26=60`

Hence, perimeter of ΔXYZ is 60 units.